# SR Problem

Discussion in 'Physics & Math' started by Aer, Aug 6, 2005.

1. ### AerRegistered Senior Member

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OK - Pete and Funkstar, here is the SR problem you wanted. Assume all time units in your favorite measurement: years, hours, minutes, etc but then distance units must be give the corresponding equivalent, light-years, light-hours, light-minutes, etc such that c=1.

We have two mother ships S1 and S2 that exist at the positions, (x,y), of (0,0) and (0,1) respectively. That is:

S1 has position x=0, y=0
S2 has position x=0, y=1

Each mother ship, S1 and S2, has a scout ship, S1' and S2' respectively. For the entire problem, we will assume S1 and S1' read each other's clocks and S2 and S2' read each other's clocks.

At t=0, each scout ship exists within their respective motherships. That is they are at rest with respect to their motherships at the position x=0. furthermore, we specify that the motherships are at rest with repect to each other. In other words, all ships, S1, S1', S2, and S2' exist in the frame S at x=0 and t=0 so all of their clocks read 0.

At t=0+ (immediately after t=0), the scout ships instantaneously accelerate to .9c with respect to the motherships which are still in frame S. The frame in which the scout ships are in after this instantaneous acceleration will be refered to as frame S'.

Now, assume the scout ships travel forever at .9c. Since the motion of the two scout ships are completely in sync, (question 1

will their clocks also be completely in sync?

At a time interval, &Delta;t=100 as measured in the S frame, the mother ship S1 instantaneously accelerates to the S' frame. That is, at t=100 in the S frame, S1 accelerates to .9c with respect to the S frame. (question 2

What is the time that S1 reads on the clock of S1'?

At a time interval, &Delta;t=10 as measured in the S frame, the mother ship S1 instantaneously decelerates back to the S frame. That is, at t=110 in the S frame, S1 decelerates to 0c with respect to the S frame. (question 3

What is the time on the clock of S1?

At a time interval, &Delta;t=10 as measured in the S frame, the scout ship S2' instantaneously decelerates back to the S frame. That is, at t=120 in the S frame, S2' decelerates to 0c with respect to the S frame. (question 4

What is the time that S2 reads on the clock of S2'?

(question 5

What is the time on the clock of S2?

Last edited: Aug 6, 2005

3. ### James RJust this guy, you know?Staff Member

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This is not an SR problem, since it involves many accelerations.

5. ### AerRegistered Senior Member

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Au contraire! See the relativistic rocket equations, I figured you of all people would know accelerations are acceptable in SR - this is even a given in the -superluminal experiment-. How do you explain the twin paradox in SR without acceleration? -the same acceleration is used here as in the twin paradox example- -instantaneous acceleration-

7. ### James RJust this guy, you know?Staff Member

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Ok, Aer. SR can handle accelerations, but if you want quantitative results you'll need to specify details of the accelerations, and set up the relevant integrations. Specifying "instantaneous acceleration" is non-physical and insufficient if you want quantitative answers.

8. ### AerRegistered Senior Member

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No it is not, the acceleration time interval is insignificant to the final result. I originally posted a similar problem to this on physicsforums but included time intervals of &Delta;t=1 for the acceleration phases. The people there complained and whined until I changed it to instantaneous acceleration (i.e. &Delta;t -> 0).

9. ### AerRegistered Senior Member

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Where are Pete and Funkstar?
The answers to these questions are very easy to compute. There is followup analysis to come

10. ### funkstarratsknufValued Senior Member

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Ok, Aer. I only saw your PM today, and though I don't quite remember what this is for...
Assuming that they are travelling along the x-axis: Yes, from both S and S'.
Assuming that they are travelling along the y-axis: No, but I need to know in what frame you consider the motion to be completely in sync from, to tell you the details of which frame they are in sync in, and which frame they are out of sync in.

I'll assume the direction of motion is the x-axis from here on (as is standard.)
&gamma; = 2.294

In the S frame, S1' is at (assuming lightseconds as units for length)

x=90
t=100

which transforms to

x'=0
t'=43.59

So, the reading is ca. 43.59s.
Okay, I'm a bit tired so I'll be lazy from now on. Depending on your reply I may do the full transforms later, but I simply don't have the energy right now.

Time dilation gives that during 10s in S, S1 will accumulate only ca. 4.36s in S', so the clock should read 94.36s.
120/&gamma; = 52.31s
I assume you mean at time t=120 in the S frame. It reads 120 second, of course.

11. ### AerRegistered Senior Member

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Regarding question 1, yes all motion is in the x direction as is standard. The y component was only mentioned to say that the two motherships did not exist in the exact same volume of space.
1) Correct.

2) Incorrect.
S' is the rest frame here since S1 is accelerating to the S' frame. so: v=.9c t=100 x=0 tp = &gamma;(t - v*x) = &gamma; * t = 2.294 * 100 = 229.4 So, time on S1' is: 229.4
3) Close, but incorrect.

4) Correct.

5) Correct.

Last edited: Aug 8, 2005
12. ### AerRegistered Senior Member

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hint: You cannot use the original lorentz setup for all answers. You must setup new lorentz transforms after each acceleration - I was hinting to this in the story problem by giving you &Delta;t values rather than just strict time measurments in the S frame. I hope this makes sense without further elaboration.

13. ### funkstarratsknufValued Senior Member

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Told you I was tired.
You're right. The frame change happens for the mothership S1, so

x=0
t=100

translates to

x'=-206.47
t'=229.42
I had gotten 90s instead 100s into it. Also, I was confusing frames. Properly, this time:

S1 will be at x = 9, t=110

translates into S' as

x'=-206.47 (good enough)
t'=233.77

So from S1's point of view (233.77-229.42) = 4.35s have passed in the S' frame + the 100 seconds before he entered it. So 104.35s (ca.)

So, what's up?

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2) Correct.

3) Correct.

15. ### AerRegistered Senior Member

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Let's look at things from the perspective of S1.

Let the first time interval, &Delta;t=100 in which S1 is stationary in the S frame be refered to as "time interval 1".

Let the second time interval &Delta;t=10 in which S1 is stationary in the S' frame be refered to as "time interval 2".

Let the third time interval &Delta;t=10 in which S1 is stationary in the S frame be refered to as "time interval 3".

From our conclusions above, we can infer that S2' is not in the S' frame according to S1 during time interval 2. Also, we can infer that S2' is not in the S frame according to S1 during the time interval 3.

But we know that S2' is either in the S' frame or the S frame. Specifically we know that S2' is in the S' frame for all t=0+ until t=120, where S2' enters the S frame. But S1 doesn't see it in either the frame S' when S1 enters S' or the frame S when S1 returns to the S frame, until t=114.4 according to the clock of S1. The relativity of simultaneity is truely beautiful.

16. ### funkstarratsknufValued Senior Member

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I'm annoyed at myself, now. I've been preaching how the Lorentz transforms is the way to do str, and the first time afterwards, I skip over them and make gross errors.

Stupid!

Oh, and I'm off to bed in a few. I'll be interested to see what you get from this.

17. ### funkstarratsknufValued Senior Member

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I must admit, I fail to see a problem.

Remember that the two frames do not agree on what those time intervals mean.

I'll sleep on it, though.

18. ### AerRegistered Senior Member

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I must admit - I thought you wouldn't see a problem as you don't accept common sense as having any say-so in Relativity. :m: :bugeye:

Would you say that S1 sees S2' at rest in the S' frame during time interval 1, at rest in the S frame during time interval 2, at rest in the S' frame during time interval 3, and at rest in the S frame at any time after as recorded by S1?

Edited to include the phrases "at rest" where appropriate.

Last edited: Aug 8, 2005
19. ### James RJust this guy, you know?Staff Member

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Objects don't exist in only one frame. A frame is a point of view. All objects exist in all frames at all times. Events in spacetime can be viewed from any frame.

Statements like "S1 moves from frame S to frame S'" are meaningless, since S1 exists in both frames always (though with different velocities).

20. ### AerRegistered Senior Member

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Of course not. That is why I refer to what frame S1 would see S2' in.

YES.

You are missing the point - by "accelerate to frame S' " I am implying the object is not at rest in S' before the acceleration and is at rest in the frame S' after the acceleration.

Good observation. Agreed.

You very well know the context of that statement is: "S1 moves from being at rest in frame S to at rest in frame S' ". That is not meaningless! Did you even read the story problem?

Wow. You are truely grasping at straws here. Try to do a little something called reading comprehension.

Funkstar was able to comprehend the story problem and come up with the correct answers. Why can't you?

Just for you, I will repeat the line in which frame S and frame S' are defined:

At t=0+ (immediately after t=0), the scout ships instantaneously accelerate to .9c with respect to the motherships which are still in frame S. The frame in which the scout ships are in after this instantaneous acceleration will be refered to as frame S'.

21. ### superluminalI am MalcomRValued Senior Member

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Sorry to pop in like this but I'm in a foul mood.

JamesR, you are clearly an idiot without the ability to read at an even gradeschool level. Your physics knowledge is that of a caveman. You have been shown to be incorrect by MacM, Aer, Geist, and many others. When will you admit you are out of your depth here?

You grasp at straws like a drowning man. Give up and let the real physics gurus here get on with things. Quit disturbing our threads with "mainstream" so-called "physics". Bah. Fool. Dupe of the establishment.

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